S1 CSE Engineering Graphics · Projection of Solids

Orthographic Projections of Solids Inclined to Both Reference Planes

This notebook teaches how to draw multi-stage orthographic projections of three-dimensional solids—prisms, pyramids, cylinders, cones, and cubes—inclined to both the Horizontal Plane (HP) and Vertical Plane (VP). Learners will master the three-stage change-of-position method, visibility rules, apparent angle (locus) constructions, and special geometric conditions such as freely suspended solids.

Before you start: First-angle orthographic projection fundamentals (XY reference line, Front View above XY, Top View below XY); Orthographic projections of points, straight lines (including true length and apparent inclination), and regular plane figures; Basic geometric constructions of regular polygons (triangles, squares, pentagons, hexagons) and division of circles.

Your path through this module

Classification of Solids and the Simple Position: Every multi-stage solid projection begins in a 'simple position' with the axis perpendicular to a reference plane, oriented so the resting edge or corner lies at the extreme right or left.
The Three-Stage Change-of-Position Method and Visibility Rules: By tilting the solid in two successive stages—first changing its elevation angle without altering widths from the XY line, then rotating its plan without altering heights—we can project any 3D orientation onto 2D paper.
Solids of Revolution and Resting on Faces, Slant Edges, or Generators: Curved solids are projected by tracking eight equally spaced generators around their circular base, and conditions like 'resting on a slant face or generator' simply mean rotating that line directly onto the XY line.
True vs. Apparent Inclination (The Locus Method) and Profile Plane Cases: When a line is already tilted to HP, its top view is foreshortened, so to incline it at a true angle to VP we must use a locus construction to find the larger apparent angle (beta) in the top view.
Freely Suspended Solids and Solids Resting on the Vertical Plane (VP): A freely hanging solid tilts until the line joining its suspension corner and its center of gravity is vertical, while a solid resting on VP simply swaps the roles of the front and top views across all three stages.
Practice: Polyhedra on HP with Direct Stage-3 Rotation: When a resting base edge lies on HP (preserving its true length) or the top-view angle with XY is explicitly given, Stage 3 is drawn by rotating the Stage 2 top view directly by the stated angle.
Practice: Cones and Cylinders on HP with Direct Stage-3 Rotation: By dividing the circular base into eight points and tracking them through three stages, we can accurately plot the elliptical projections of cones and cylinders tilted on base points or generators.
Practice: Pyramids and Cubes on HP with Special Face, Edge, and Diagonal Conditions: Geometric statements like 'resting on a triangular face', 'slant edge vertical', or 'body diagonal perpendicular to VP' replace numeric angles by telling us which line in the solid must be made horizontal or vertical.
Practice: True Inclination to Both Planes (Locus and Profile Plane Problems on HP): Whenever a problem specifies the true angle of an already-inclined axis or slant edge with VP, you must first construct a locus triangle to find the enlarged apparent angle beta before rotating the Stage 2 top view.
Practice: Solids Resting on the Vertical Plane (VP): When a solid rests on VP, the entire three-stage procedure is mirrored across the XY line: draw the true base in the Stage 1 front view, tilt to VP in the Stage 2 top view, and rotate to HP in the Stage 3 front view.
Practice: Real-World Equilibrium—Freely Suspended Pyramids: A pyramid suspended from a base corner naturally settles so that its center of gravity (one-quarter of the height up the axis) hangs directly below the suspension corner.

Classification of Solids and the Simple Position

You already know how to project points, lines, and flat planes onto the Horizontal Plane (HP) and Vertical Plane (VP). This lesson shows you how to set up three-dimensional solids in their starting upright view—the essential first step before tilting them in space.

Look around a workshop and you will see two main families of machine parts: flat-sided blocks and round turned shafts. In engineering drawing, flat-sided solids are called polyhedra, and round ones are solids of revolution. Both share an axis, which is an imaginary center line running from the center of the bottom base to the top.

Solid Type	Base & Top	Side Surfaces	Key Lines
Regular Prism	Two identical polygons	Rectangular lateral faces	Parallel longer edges
Cube (Hexahedron)	Six equal squares	Square faces	12 equal edges
Regular Pyramid	One polygon base, one point (apex)	Triangular lateral faces	Slant edges meeting at the apex
Cylinder / Cone	Two circles / One circle and an apex	Curved surface	Imaginary straight generators

Because a cylinder or cone has no sharp side edges, we divide its base circle into 8 or 12 equal parts and draw straight lines—generators—up the curved surface to give us points to track.

You cannot draw a tilted solid directly because neither the top view nor the front view shows its true shape. Instead, we start in a simple position, standing the solid upright so its base is flat against one reference plane. The rule is: if the solid will first tilt relative to the HP, stand it with its axis perpendicular to the HP; if it will first tilt relative to the VP, place its axis perpendicular to the VP. On your drawing, the view looking straight down the axis shows the base in its exact true size and shape.

Before drawing that true-shape polygon, imagine tipping a cereal box onto one bottom edge versus tipping it onto a single sharp corner. The part that stays on the table acts as a hinge. In your first view, you must place this hinge at the extreme right (or extreme left). If the solid rests on a base edge, draw that edge perpendicular to the reference line $XY$ on the right side. If it rests on a base corner, place that single point at the far right, with the two meeting edges angled equally away from $XY$ .

Worked example

Set up the simple position for a pentagonal pyramid (base side 30 mm, axis 60 mm) that will rest on a base edge on the HP and tilt toward the HP.

Choose the axis plane: Since the first tilt is with the HP, place the axis perpendicular to the HP. The top view will show the true pentagonal base.
Orient the base (Top View): Below the $XY$ line, draw a vertical line 30 mm long on the right side. This is the resting base edge, perpendicular to $XY$ . Complete the regular pentagon to the left of this edge, and connect all five corners to the center to show the slant edges meeting at the apex.
Project the elevation (Front View): Project the corners up to the $XY$ line to form the flat base. From the center, measure 60 mm straight up along the axis to mark the apex, and connect it to the base corners to finish the triangle.

Remember

Tilt with HP first $\rightarrow$ start with the axis perpendicular to HP (true base in the top view).
Resting on a base edge $\rightarrow$ draw that edge perpendicular to $XY$ on the extreme right or left.
Resting on a base corner $\rightarrow$ place that single corner on the extreme right or left.

The Three-Stage Change-of-Position Method and Visibility Rules

In the previous lesson, you learned to draw a solid standing upright in its simple position with the resting edge or corner placed at the extreme right or left. Now, we will use a step-by-step graphical method to tilt that solid to one reference plane and rotate it to the second, while tracking which edges remain visible to the observer.

Imagine tilting a hardcover book on your desk and then spinning it. Trying to draw that final, doubly tilted book all at once is nearly impossible because every edge appears foreshortened. Instead, we use the three-stage change-of-position method, a technique that breaks a complex 3D movement into three easy 2D drawings: Stage 1 is the simple position, Stage 2 tilts the solid to one plane, and Stage 3 rotates it to the second plane.

In Stage 2, we tilt the Stage 1 front view (or top view) without changing its shape. Because the axis is perpendicular to the base, tilting the axis to the Horizontal Plane (HP) by an angle $\theta_{\text{axis}}$ automatically tilts the base by the complementary angle: $\theta_{\text{base}} = 90^\circ - \theta_{\text{axis}}$ To build the new matching view, we use the projector intersection rule: draw vertical projectors (straight construction lines) from the tilted view and intersect them with horizontal locus lines (paths of constant distance from the $XY$ line) coming from the Stage 1 view. In simple words, when a solid tilts up or down in the front view, its corners keep their exact front-to-back distances in the top view.

In Stage 3, we take the newly drawn Stage 2 top view and rotate it as a rigid shape so its axis makes the required angle with the Vertical Plane (VP). Watch the directional instruction carefully: place the apex (top point) closer to the $XY$ line if it is "nearer to VP," or farther down if it is "nearer to the observer." Projecting vertically up to meet horizontal locus lines from the Stage 2 front view gives the final Stage 3 front view.

Finally, a solid is opaque, so you must distinguish continuous visible lines from dashed hidden edges (edges blocked from sight by the body of the solid). Always apply four visibility rules in order:

The outer boundary (outline) of any view is always a continuous visible line.
Look at the adjacent view to find the corner closest to the observer; all edges meeting at this nearest corner are visible.
Find the corner farthest from the observer (closest to the $XY$ line when looking from the front or top); all internal edges meeting at this hidden corner are dashed.
Two continuous visible lines can never cross each other inside the boundary—if two lines intersect, at least one must be dashed.

Worked example

A square pyramid (base 30 mm, axis 50 mm) rests on a base edge on HP with its axis inclined at 40° to HP, and the resting edge inclined at 30° to VP.

Draw Stage 1 (Simple Position): Draw a 30 mm square below $XY$ with its right edge perpendicular to $XY$ (so the resting edge is on the extreme right). Project up to draw the triangle front view of height 50 mm.
Tilt for Stage 2 (Inclined to HP): Since $\theta_{\text{axis}} = 40^\circ$ , calculate $\theta_{\text{base}} = 90^\circ - 40^\circ = 50^\circ$ . Redraw the Stage 1 triangle along $XY$ with its base tilted at 50° and the right base corner touching $XY$ .
Project the Stage 2 top view: Drop vertical projectors from the tilted triangle to meet horizontal locus lines from the Stage 1 square. Connect the outer boundary continuously. In the tilted front view, the resting right edge is lowest (farthest from a top observer), so draw that edge as a dashed line in the Stage 2 top view.
Rotate for Stage 3 (Inclined to VP): Copy the entire Stage 2 top view, rotating it so the dashed resting edge makes 30° with $XY$ . Project vertically up to meet horizontal locus lines from the Stage 2 front view, and apply the four visibility rules to finish.

Remember

Never change the shape or size of the view you are rotating between stages; only change its angle.
Base and axis inclinations to the same plane always add up to 90° ( $\theta_{\text{base}} = 90^\circ - \theta_{\text{axis}}$ ).
Always draw the outer boundary continuous first, then check the adjacent view to find the nearest (visible) and farthest (hidden) corners.

Solids of Revolution and Resting on Faces, Slant Edges, or Generators

You already know how to tilt prisms and pyramids through given angles using the three-stage method. This lesson shows you how to project curved solids that have no sharp corners, and how to tilt a solid when the problem gives a physical resting condition instead of a number of degrees.

Imagine trying to project a smooth traffic cone onto paper; unlike a pyramid, it has no sharp slant edges to track from view to view. To give ourselves points to project, we invent imaginary straight lines on the curved surface called generators, which run from the circular base to the apex (for a cone) or between the two circular bases (for a cylinder). In Stage 1, divide the circular view into 8 equal parts ( $45^\circ$ sectors) using your $45^\circ$ set-square, label the 8 base points, and project them to the other view as light generator lines.

When you tilt a cone or cylinder in Stage 2 and project those 8 base points, the circle foreshortens into an oval shape called an ellipse. Join the 8 projected points with a smooth freehand curve. To finish the solid's outer boundary, draw straight lines from the apex (or second base) that just kiss the outside of the ellipse as tangents—do not simply stop at the widest numbered points.

Sometimes a problem does not state an angle, such as tipping a party hat until its side lies flat on a table, or standing a pyramid so one sloping side is vertical. Because your Stage 1 right-hand line represents the true length of a slant edge, triangular face, or extreme generator, you can rotate the Stage 1 figure directly to match the words of the problem:

Problem Condition	Stage 2 Action (Front View)
Resting on a slant edge, triangular face, or generator on HP	Rotate the Stage 1 front view until that right-hand line lies flat on the $XY$ line.
Slant edge or triangular face perpendicular to HP	Rotate the Stage 1 front view until that right-hand line stands vertically ( $90^\circ$ to $XY$ ).

A cube has a special internal line called the body diagonal (or space diagonal), which connects a bottom corner to the opposite top corner straight through the center of the block. If you place the cube in Stage 1 with a base diagonal parallel to $XY$ , the front view shows the true length of this body diagonal as a slanted line from the bottom-left to the top-right corner. You can then rotate the Stage 2 front view to make this diagonal either parallel or perpendicular to $XY$ .

Worked example

Draw a cone (base diameter 40 mm, axis 60 mm) resting on a generator on HP, with its axis parallel to VP.

Draw Stage 1 in simple position: Draw a circle of diameter 40 mm in the top view below $XY$ . Divide it into 8 equal parts, labeling the horizontal diameter $a$ - $e$ (with $e$ on the extreme right). Project these up to the $XY$ line, mark the apex $o'$ 60 mm above the center, and draw the triangular front view. The right-hand boundary $o'e'$ is the extreme generator.
Tilt onto the generator in Stage 2: Copy the triangular front view so that line $o'e'$ lies completely flat along the $XY$ line. Use your compass to transfer the base width $a'e'$ and side $o'a'$ so the triangle stays rigid.
Project the Stage 2 top view: Drop vertical projectors from the tilted apex $o'_1$ and all 8 base points on $a'_1e'_1$ , intersecting horizontal projectors from the Stage 1 circle. Join the 8 base intersections into a smooth ellipse, and draw two straight lines from the new apex $o_1$ tangent to the outer curve of the ellipse.

Remember

Divide circular bases into 8 equal sectors in Stage 1 so you have points to project as ellipses in later stages.
Always draw the outer boundaries of tilted cones and cylinders tangent to the smooth ellipse.
"Resting on HP" means rotating the Stage 1 right-hand edge, face, or generator directly onto the $XY$ line in Stage 2.

True vs. Apparent Inclination (The Locus Method) and Profile Plane Cases

So far, we have rotated the second-stage view directly by the angle given in the problem, which only works when that angle is the view's apparent angle or the tilted line is still in true length. This lesson teaches you how to handle problems where an already tilted axis or edge is given a true angle to the second reference plane.

Imagine holding a pencil tilted up from your desk (the Horizontal Plane, HP) and ALSO angled toward the wall behind it (the Vertical Plane, VP). Because the pencil slopes upward, its shadow on the desk is shorter than the actual pencil; this shrinking in a projection is called foreshortening. If you now pivot the pencil without changing its upward tilt, that shorter shadow makes a steeper angle with the wall line than the real pencil makes in 3D space. We call the real 3D angle the true inclination ( $\phi$ ) and the steeper angle of the shadow on the drawing the apparent inclination ( $\beta$ ).

Look closely at the wording in your problem statement to know whether to rotate directly or find $\beta$ first:

Problem Wording	Angle Given	Stage 3 Action
"Top view of axis inclined at $\phi$ to VP" or "vertical plane containing axis inclined at $\phi$ "	Apparent angle ( $\beta$ )	Rotate Stage 2 plan directly at $\phi$ to XY.
"Axis (or slant edge) is inclined at $\theta$ to HP and $\phi$ to VP"	True angle ( $\phi$ )	Construct the locus to find $\beta$ ( $\beta > \phi$ ).

A locus is simply a line showing all possible positions of a moving point at a constant distance from a reference plane. When the axis has true length $L$ in Stage 1 and foreshortened length $L_1$ in the Stage 2 top view, you cannot place $L_1$ directly at the true angle $\phi$ to XY. Instead, you first draw $L$ at $\phi$ to find how far the end of the axis sits from XY (its locus), and then fit the shorter length $L_1$ into that exact same distance. On your drawing, the shorter line $L_1$ must swing steeper to reach the same locus line, automatically giving you the apparent angle $\beta$ .

A special case occurs when the two true angles add up to 90° ( $\theta + \phi = 90^\circ$ ). Hold your pencil up at 30° to the floor and 60° to the wall: it must point straight toward the wall, lying entirely in a Profile Plane (PP)—a flat plane perpendicular to both HP and VP. Whenever $\theta + \phi = 90^\circ$ , the apparent angle $\beta$ is simply 90°, so you rotate the Stage 2 top view until the axis is perpendicular to the XY line without needing a locus construction.

Worked example

Suppose a cone's axis of true length $L = 60$ mm is inclined at $\theta = 30^\circ$ to HP and $\phi = 45^\circ$ to VP. After drawing Stages 1 and 2, the Stage 2 top view of the axis measures a foreshortened length $L_1 = 52$ mm. Let us find the Stage 3 orientation:

Draw the true angle line: Below the XY line, pick a starting point $o_1$ (representing the apex) and draw a light construction line inclined at the true angle $\phi = 45^\circ$ to XY. This models the true 3D tilt before foreshortening.
Establish the locus: Along this 45° line, measure the true axis length $L = 60$ mm from $o_1$ to mark point $p$ , and draw a horizontal line (parallel to XY) through $p$ . This horizontal line is the locus of the base center, fixing its true distance in front of VP.
Find the apparent angle ( $\beta$ ): With your compass set to the foreshortened Stage 2 axis length $L_1 = 52$ mm and centered at $o_1$ , swing an arc to cut the horizontal locus line at $p_1$ . Join $o_1$ to $p_1$ ; this line makes the steeper apparent angle $\beta$ (about 55°) with XY, and you now redraw the Stage 2 top view along $o_1 p_1$ .

Remember

Only use the locus construction when the line being angled to VP is already inclined to HP (so its top view is foreshortened) AND the problem gives its true angle to VP.
The apparent angle $\beta$ is always larger than the true angle $\phi$ because the foreshortened length $L_1$ is shorter than the true length $L$ .
If the true angles with HP and VP add up to 90° ( $\theta + \phi = 90^\circ$ ), skip the locus construction and place the Stage 3 axis line perpendicular (90°) to the XY line.

Freely Suspended Solids and Solids Resting on the Vertical Plane (VP)

You have learned how to tilt solids by given angles and find apparent angles using the locus method. This final theory lesson shows you how to draw a solid hanging freely from a string, and how to handle solids that rest against the vertical wall (VP) instead of the floor (HP).

Imagine tying a string to one corner of a heavy wooden pyramid and lifting it into the air. The pyramid will swing and settle into a natural tilt. This happens because of its center of gravity (CG), the single balance point where all the solid's weight acts. For any prism or cylinder, the CG is on the axis at half the height ( $h/2$ ). For any pyramid or cone, because most of the bulk is near the bottom, the CG lies on the axis at one-fourth the height ( $h/4$ ) measured up from the base.

When a solid hangs freely from a corner, gravity pulls the CG straight down until it rests directly below the hanging corner. The straight line joining the suspension corner to the CG is called the line of suspension. On your drawing, you mark the CG as $g'$ on the axis in the Stage 1 front view, join it to the hanging corner (placed at the extreme right or left), and then rotate the Stage 2 front view so this line of suspension stands perpendicular to the $XY$ line, with the corner at the top.

Now imagine pushing a solid against a wall (VP) instead of setting it on the floor (HP). Everything you learned about the three stages still works, except the front view and top view swap jobs. You start Stage 1 with the axis perpendicular to VP, so the true shape of the base appears in the front view.

Stage	Solid Resting on HP	Solid Resting on VP
Stage 1	Top view shows true base.	Front view shows true base.
Stage 2	Rotate front view (tilt to HP).	Rotate top view (tilt to VP).
Stage 3	Rotate top view (tilt to VP).	Rotate front view (tilt to HP).

Worked example

A square pyramid (base 30 mm, axis 60 mm) is freely suspended from one of its base corners so that its axis is parallel to VP. Draw its projections.

Draw Stage 1 on HP: Since a corner is the key point, draw a 30 mm square in the top view with all sides at 45° to $XY$ . Name the extreme right corner $c$ .
Project the Stage 1 front view: Draw the triangle of height 60 mm sitting on $XY$ . The extreme right base corner is $c'$ .
Locate the CG: Measure $60 / 4 = 15$ mm up along the axis from the base and mark $g'$ . Draw a light line joining $c'$ to $g'$ .
Rotate for Stage 2: Copy the front view triangle Above $XY$ so that the line $c'g'$ is perpendicular to $XY$ , with $c'$ above $g'$ . Project down from Stage 2 and across from Stage 1 to complete the top view.

Remember

The center of gravity (CG) of a pyramid or cone is on its axis at $h/4$ from the base; for a prism or cylinder, it is at $h/2$ .
For a freely suspended solid, rotate the Stage 2 front view until the line joining the hanging corner and $g'$ is perpendicular to $XY$ .
When a solid rests on VP, start with the true base in the front view, tilt the top view in Stage 2, and rotate the front view in Stage 3.

Practice: Polyhedra on HP with Direct Stage-3 Rotation

With the three-stage method and visibility rules in hand, we start practicing on prisms and pyramids resting on HP. What is new here is recognizing when the third stage requires a direct angle rotation without a locus construction, and controlling the exact quadrant orientation of the solid.

The idea

In a three-stage projection, Stage 3 rotates the Stage 2 top view without changing any point's height above HP. You do not need a locus construction whenever the line whose inclination to VP is given appears in true length in the Stage 2 top view, or when the problem directly states the angle made by the top view with the $xy$ line.

First, consider a solid resting on a base edge on HP. Because that resting edge lies flat on HP throughout Stage 2, its top view remains in true length. When the problem states that this resting base edge is inclined at an angle $\phi$ to VP, you simply rotate the Stage 2 top view so that the resting edge makes the true angle $\phi$ directly with the $xy$ line. Similarly, if the problem states that the vertical plane containing the axis (or the top view of the axis or slant edge) is inclined at $\phi$ to VP or $xy$ , that angle is already a plan angle, so you rotate the top view directly by $\phi$ .

Second, pay attention to apex orientation in pyramids. In Stage 2, tilting the solid on a right-side base edge tips the apex to the right; tilting it on a left-side base edge tips the apex to the left. In Stage 3, rotating the top view so the apex moves below the resting edge (further from $xy$ ) places the apex away from VP, whereas rotating it so the apex lies closer to $xy$ places the apex towards VP.

Worked example

Problem Q211 (51): Triangular prism resting on a base edge inclined to VP Given: A triangular prism of base edge 30 mm and length 50 mm rests on a base edge on HP, with its axis inclined at 35° to HP and the resting base edge inclined at 45° to VP. Find: Draw the three-stage orthographic projections of the prism.

Stage 1 (Simple position): Draw the top view as an equilateral triangle $abc(123)$ of side 30 mm below $xy$ with edge $b(2)-c(3)$ vertical on the right so the resting edge becomes a point in elevation; project the front view as a rectangle of height 50 mm above $xy$ .

Figure: Step 211-1: Initial projection of triangular prism resting with its base on HP

Initial stage (Step 211-1) of projecting a triangular prism with base edge 30 mm and height 50 mm:

A horizontal reference line labelled 'x' on the left and 'y' on the right.

Text boxes to the left of the drawing specify: 'B = 30', 'H = 50', 'SCALE 1:1', and 'ALL DIMENSIONS ARE IN mm'.

Top view (below xy line): An equilateral triangle with edge length 30 mm, oriented such that vertical edge c(3)-b(2) is perpendicular to the xy line on the right, and apex a(1) points to the left. Dimension 30 is shown along the edge a(1)-b(2).

Front view (above xy line): A rectangle of height 50 mm resting on the xy line. The bottom corners are labelled 1' on the left and 2'(3') on the right. The top corners are labelled a' on the left and b'(c') on the right. The left vertical edge has a height dimension of 50.

Vertical projection lines connect top view vertex a(1) to front view base point 1' (and top a'), and edge c(3)-b(2) to base point 2'(3') (and top b'(c')).

Stage 2 (Front view): Redraw the front view tilted on the right base edge $2'(3')$ on $xy$ so the axis is inclined at 35° to $xy$ (the base makes $90^\circ - 35^\circ = 55^\circ$ with $xy$ ).

Figure: Orthographic projection of a triangular prism - Step 211-2: Front view tilted at 35 degrees

Step 211-2 of triangular prism projection. Given base edge B = 30 mm, height H = 50 mm. In the first position: Top view shows an equilateral triangle with vertices a(1) on the left, b(2) and c(3) on a vertical line on the right; edge a(1)-b(2) is dimensioned as 30. Front view directly above XY line is a vertical rectangle of height 50 mm and base spanning from projector 1' to projector 2'(3'), top vertices are a' and b'(c'). In the second position: Front view is rotated and rests on point 2'(3') on the XY line, with its bottom face inclined at an angle of 35° to the XY line; vertices labelled 1', 2'(3'), a', and b'(c'). The second top view has not yet been projected.

Stage 2 (Top view): Project vertically down from the tilted front view and horizontally from the Stage 1 top view to complete the foreshortened Stage 2 top view, marking the hidden lower edge $2-3$ as dashed.

Figure: Orthographic projection of a triangular prism - Step 211-3: Projection of final top view

Step 211-3 of triangular prism projection. Given base edge B = 30 mm, height H = 50 mm. Stage 1 shows the reference front view (rectangle of height 50 mm) and top view (triangle with side 30 mm). Stage 2 shows the front view inclined at 35° resting on base corner 2'(3') on the XY line. In Stage 3, projectors are dropped down from the tilted front view and drawn horizontally from the initial top view to construct the second top view: visible outline formed by points 1, 2, b, c, 3, 1; vertex a inside connects to b and c with solid lines; hidden edge connecting base vertices 2 and 3 is shown as a dashed line.

Stage 3 (Top view): Because the resting base edge $2-3$ stays on HP in true length, rotate the Stage 2 top view directly so edge $2-3$ is inclined at 45° to $xy$ .

Figure: Orthographic projections of a triangular prism in three stages (Problem 211-4)

Three-stage projection of an equilateral triangular prism with base side B = 30 mm and axis height H = 50 mm. Scale 1:1, all dimensions in mm. Given box: B = 30, H = 50. Stage 1 (Simple position): Front view is a rectangle of height 50 mm resting on the XY line, labeled a' and b'(c') at the top, 1' and 2'(3') on the XY line. Top view is an equilateral triangle of side 30 mm, vertex a(1) pointing to the left and vertical side b(2)-c(3) on the right. Stage 2 (Base/axis inclined to HP): Front view is tilted such that the base edge 2'(3') rests on the XY line, inclined at 35° to the horizontal. Vertices are labeled 1', a', b'(c'), 2'(3'). Top view is projected downwards from front view and horizontally from stage 1 top view; hidden edge 2-3 is shown dashed, vertices labeled 1, 2, 3 and a, b, c. Stage 3 (Inclined to VP): Top view is reproduced with the axis/edge inclined at 45° to the projector line, vertices labeled 1, 2, 3 and a, b, c with edge 2-3 dashed.

Stage 3 (Front view): Project vertically up from the Stage 3 top view and horizontally from the Stage 2 front view to complete the final front view, drawing visible edges solid and hidden edges dashed.

Figure: Complete orthographic projections of a triangular prism in three stages (Problem 211-5)

Complete three-stage orthographic projection of an equilateral triangular prism with base side B = 30 mm and height H = 50 mm. Scale 1:1, all dimensions in mm. Given box: B = 30, H = 50. Stage 1: Front view is a rectangle with height 50 mm resting on the reference line x-y, labeled a', b'(c') on top and 1', 2'(3') at bottom. Top view is an equilateral triangle with side 30 mm, vertices a(1) pointing left and c(3)-b(2) on the right. Stage 2: Front view is tilted with base edge 2'(3') on the reference line, inclined at 35° to the HP; vertices labeled 1', a', b'(c'), 2'(3'). Top view projected below with hidden edge 2-3 dashed; vertices 1, 2, 3 and a, b, c. Stage 3: Top view reproduced inclined at 45° to the vertical projector line; vertices 1, 2, 3 and a, b, c with hidden edge 2-3 dashed. Final front view projected vertically from stage 3 top view and horizontally from stage 2 front view, showing all visible and hidden edges connecting vertices 1', 2', 3' and a', b', c'.

Answer: The three-stage projections of the 30 mm by 50 mm triangular prism with axis at 35° to HP and resting edge at 45° to VP are completed as shown.

Practice: Cones and Cylinders on HP with Direct Stage-3 Rotation

Having practiced direct three-stage rotations on flat-faced polyhedra, we now apply the exact same direct rotation method to curved solids—cones and cylinders—using eight generators. By dividing the circular base into eight points and tracking them through three stages, we can accurately plot the elliptical projections of cones and cylinders tilted on base points or generators.

The idea

Unlike prisms and pyramids, cones and cylinders have smooth curved surfaces and circular bases with no sharp corners to project. To track them across three stages, we divide the circular base in the Stage 1 top view into eight equal parts (every 45°), numbered 1 to 8, and draw the corresponding generators—straight lines on the curved surface running from the base points to the cone's apex (or between the cylinder's two bases). When the solid tilts in Stage 2 and rotates in Stage 3, the circular base foreshortens into an ellipse. Joining the eight projected base points with a smooth freehand curve gives that ellipse, and drawing tangents from the apex (or between the two ellipses) gives the outer visible boundary of the curved surface.

In every problem here, Stage 3 requires a direct rotation of the Stage 2 top view without a locus construction. This happens for one of three reasons: the top-view angle with the XY line is given directly (or as the vertical plane containing the axis and resting generator), the angle is given for the base diameter parallel to HP (chord 3–7, which stays perpendicular to XY in Stage 2 and therefore keeps its true length), or a geometric condition fixes the Stage 3 top view directly (such as the axis lying on a Profile Plane perpendicular to XY, or a generator lying directly on XY).

When finishing each view, always check base visibility first: if the base is tilted toward the observer (or toward the top in Stage 2), the entire elliptical base is a continuous dark curve; if it faces away, the inner half of the ellipse is hidden and drawn dashed while the outer boundary generators remain continuous.

Worked example

Problem 69: A cone of base diameter 40 mm and height 60 mm is resting on a base point on HP. The axis (or generator) is inclined at 40° to HP and the vertical plane containing the axis and the generator is inclined at 45° to VP with the apex away from VP.

Given: Cone with base diameter $D = 40$ mm, height $H = 60$ mm, resting on base point $5'$ on HP, tilted at 40° to HP, and vertical plane containing the axis inclined at 45° to VP (apex away from VP). Find: The three-stage orthographic projections of the cone.

Stage 1 (Simple position): Draw the top view as a circle of diameter 40 mm below XY, divide it into 8 equal parts numbered 1 to 8 with point 5 at the extreme right so it becomes the resting point, and project the front view as an isosceles triangle of base 40 mm on XY and height 60 mm.

Figure: Step 69.1: Simple position of the cone with base on HP

Orthographic projection showing the first step (labelled 69.1) of a cone in simple position resting on HP:

Parameter box specifies: 'D = 40', 'H = 60'.

Horizontal reference line labelled 'x' on the left and 'y' on the right.

Top View (below xy): Circle of diameter 40 mm with center 'o', divided into 8 equal sectors with division points numbered 1 (extreme left), 2, 3 (bottom), 4, 5 (extreme right), 6, 7 (top), 8.

Front View (above xy): Isosceles triangle standing on the xy line representing the cone of height 60 mm. The base is on xy, with base points marked 1', 2'(8'), 3'(7'), 4'(6'), 5' from left to right. The apex is labelled o'. Generators connect o' to the base points. Center line/axis extends from 3'(7') to apex o'. Vertical projector lines connect the top view circle divisions to the front view base points.

Stage 2 Front View: Copy the front-view triangle with point $5'$ resting on the XY line and incline the right-hand generator $5'o'$ at 40° to XY so the cone tilts to the right.

Figure: Orthographic projection of a cone - Step 69.2

Step 69.2 showing the tilt of the front view of a cone with base diameter D = 40 mm and height H = 60 mm.

Reference line labeled 'x' and 'y'.

Simple position:

Top view: Circle with diameter D = 40 divided into 8 equal parts labeled 1 (left), 2, 3 (bottom), 4, 5 (right), 6, 7 (top), 8, with center 'o'.

Front view: Triangle resting on HP (xy line). Base points labeled 1', 2'(8'), 3'(7'), 4'(6'), 5'. Apex labeled o' at height H = 60.

Second position front view:

The cone is rotated so that the generator 5'-o' is inclined at an angle of 40° to the xy line (horizontal).

Base labeled 1', 2'(8'), 3'(7'), 4'(6'), 5' and apex o'.

Parameters box: D = 40, H = 60.

Stage 2 Top View: Drop vertical projectors from $1' ext{ to } 5'$ and $o'$ to meet horizontal projectors from the Stage 1 circle; join points 1 to 8 with a smooth ellipse (keeping the lower half dashed if hidden from above) and draw the two extreme tangent generators to apex $o$ .

Figure: Orthographic projection of a cone - Step 69.3

Step 69.3 showing the completed second stage projection of the cone.

Simple position (left):

Front view: Triangle on xy line with apex o' and base points 1', 2'(8'), 3'(7'), 4'(6'), 5'.

Top view: Circle with 8 generator divisions (1 to 8) and center o.

Second stage (right):

Front view: Cone tilted with generator 5'-o' inclined at 40° to the xy line, resting on point 5' on HP.

Top view: Obtained by projecting points vertically down from the front view and horizontally across from the initial top view.

The circular base appears as an ellipse with points 1, 2, 3, 4, 5, 6, 7, 8. The portion of the rim from 4 through 5 to 6 is hidden and drawn with a dashed line.

Tangent generators are drawn from apex o to the elliptical base.

Parameters box: 69.3, D = 40, H = 60.

Stage 3 Top View: Because a vertical plane appears as a straight line in the top view, rotate the entire Stage 2 top view directly so the axis line $o ext{–}5$ makes 45° with the XY line, placing apex $o$ closer to XY (away from the observer, or toward VP as drawn).

Figure: Projection of a cone: step 69.4 showing stage 1, stage 2, and stage 3 top view

Step 69.4 of projection of a cone with base diameter $D = 40\text{ mm}$ and height $H = 60\text{ mm}$ . Reference line $xy$ is horizontal.

Stage 1 (Simple Position):

Top View: Circle of diameter $40\text{ mm}$ with center $o$ , divided into 8 equal sectors with generator points 1 to 8 clockwise (1 at left, 3 at bottom, 5 at right, 7 at top).

Front View: Isosceles triangle with base on $xy$ line from $1'$ to $5'$ , apex $o'$ at height $60\text{ mm}$ . Base generator points labeled $1', 2'(8'), 3'(7'), 4'(6'), 5'$ . Center axis connects $3'(7')$ to apex $o'$ .

Stage 2 (Axis inclined at $40^\circ$ to HP):

Front View: The cone from Stage 1 is tilted about the base point $5'$ on the $xy$ line such that the axis makes an angle of $40^\circ$ with the horizontal.

Top View: Projected from Stage 2 front view and Stage 1 top view. The base is an ellipse with points 1 to 8; apex $o$ is to the right. Outline generators connect $o$ to the tangents of the ellipse. The visible curve of the base is drawn with a solid line, and the hidden portion is dashed.

Stage 3 (Axis inclined at $45^\circ$ to VP - Top View only):

Top View: Redrawn from Stage 2 top view with the axis tilted at $45^\circ$ to the vertical projector direction, apex $o$ directed toward the lower right.

Stage 3 Front View: Project all eight base points and apex $o$ vertically up to meet horizontal projectors from the Stage 2 front view, draw the smooth elliptical base (with the back half dashed), and connect the extreme tangent generators to apex $o'$ .

Figure: Projection of a cone: step 69.5 showing completed projection with final front view

Step 69.5 of projection of a cone with base diameter $D = 40\text{ mm}$ and height $H = 60\text{ mm}$ , showing the completed three-stage projection including the final front view.

Stage 1 and Stage 2 are identical to step 69.4.

Stage 3 (Final views):

Top View: Same as in step 69.4, axis inclined at $45^\circ$ to VP.

Front View: Obtained by projecting vertically upwards from the Stage 3 top view and horizontally from the Stage 2 front view. Base points $1', 2', 3', 4', 5', 6', 7', 8'$ form an ellipse with hidden portions drawn dashed. Apex $o'$ is located at the top right at the same level as in Stage 2. Extreme generators connect apex $o'$ tangent to the base ellipse to form the final front view of the cone.

Watch out. The problem text states the axis is inclined at 40° to HP and the apex is away from VP, whereas the worked drawing inclines generator $5'o'$ at 40° to XY and places the apex closer to XY in the top view.

Answer: Three-stage projections of the cone ( $D = 40$ mm, $H = 60$ mm) inclined at 40° to HP and 45° to XY in the top view.

Practice: Pyramids and Cubes on HP with Special Face, Edge, and Diagonal Conditions

Now that you can tilt solids by explicit degree angles, we will practice problems where the Stage 2 tilt is defined geometrically by placing a face, slant edge, apex, or cube diagonal parallel or perpendicular to a reference plane. Instead of measuring a numeric angle, geometric statements like "resting on a triangular face", "slant edge vertical", or "body diagonal perpendicular to VP" tell us directly which line in the solid must be made horizontal or vertical.

The idea

When a problem states a geometric resting condition instead of a degree angle, the shape of the solid itself controls the tilt. In Stage 1, you must orient the base so the featured element—a triangular face (a flat lateral side of a pyramid), a slant edge (a corner line running from the base to the apex), or a body diagonal (the internal line connecting two opposite corners through the center of a cube)—appears as a single true-length line on the extreme right or left of the front view.

In Stage 2, you simply rotate the Stage 1 front view until that edge-view line matches the stated geometry. If the pyramid is lying on a triangular face or slant edge on HP, rotate that line directly onto the $xy$ line (making it horizontal). If a face or slant edge is perpendicular to HP (vertical), rotate that line to $90^\circ$ with the $xy$ line. If the pyramid rests on its apex on HP, draw the Stage 1 front view upside down with the apex $o'$ on $xy$ and the base at the top.

In Stage 3, translate the second condition into a top-view rotation. For example, making a vertical triangular face "parallel to VP" or making a horizontal body diagonal "perpendicular to VP" simply means rotating the Stage 2 top view by $90^\circ$ .

Worked example

Problem 213: A square pyramid of base edge 30 mm and height 60 mm is resting on HP on its triangular face such that the square base edge on HP is inclined at $30^\circ$ to VP. Draw its projections.

Given: Square pyramid, base edge $B = 30$ mm, height $H = 60$ mm, resting on HP on a triangular face, resting base edge inclined at $30^\circ$ to VP. Find: Three-stage front and top view projections.

Stage 1 (Simple position): Draw the top view as a square $abcd$ of side 30 mm below $xy$ with right-hand edge $bc$ perpendicular to $xy$ so the triangular face $obc$ will appear as a single edge in the front view. Join the diagonals to mark the apex $o$ . Project up to draw the front view triangle $a'(d')-b'(c')-o'$ of height 60 mm resting on $xy$ .

Figure: Step 213.1: Projections of square pyramid in simple position

Step 213.1 of the solution showing the orthographic projections of a square pyramid in simple position resting on HP with its base.

Front view (above ground line x-y): Triangle with base on the x-y line spanning from a'(d') to b'(c') and vertex o' at the top. The vertical axis height is dimensioned as 60 mm.

Top view (below ground line x-y): A square base with vertices a (bottom-left), b (bottom-right), c (top-right), and d (top-left). The base edge is dimensioned as ◻30. The two diagonal lines intersect at the apex/center o, representing the slant edges of the pyramid.

Projection lines: Vertical alignment between top view vertices (a, d) and front view base corner a'(d'), top view vertices (b, c) and front view base corner b'(c'), and center o to apex o'.

Info box labelled '213.1' with 'B = 30' and 'H = 60'.

Stage 2 Front View (Lying on triangular face): Rotate the front view triangle to the right until the right-hand slant face $b'(c')-o'$ lies flat on the $xy$ line. Use compass arcs from $b'(c')$ and $o'$ to locate the lifted base edge $a'(d')$ .

Figure: Step 213.2 of projection of a square pyramid lying on a slant triangular face or edge on the HP

Projection of a square pyramid with base side B = 30 mm and height H = 60 mm (Step 213.2).

Left (Simple position):

Reference line xy.

Front view: Isosceles triangle with base a'(d') - b'(c') = 30 mm resting on xy, apex o' at a vertical height of 60 mm.

Top view: Square of side 30 mm with corners a (bottom-left), b (bottom-right), c (top-right), d (top-left), diagonals intersecting at apex o.

Right (Tilted position):

Front view: The pyramid is tilted such that the slant edge/face b'(c')-o' lies horizontally on the xy line. Apex o' is on xy to the right, b'(c') is on xy to the left, and apex a'(d') is elevated above xy.

Labels and annotations: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm', label '213.2', and parameter box 'B = 30', 'H = 60'.

Stage 2 Top View: Project vertical lines down from $a'(d')$ , $b'(c')$ , and $o'$ in the tilted front view, and intersect them with horizontal locus lines from the Stage 1 top view. Darken the visible upper faces and draw the hidden bottom base edge $bc$ as a dashed line.

Figure: Step 213.3 of projection of a square pyramid showing top view of tilted pyramid

Projection of a square pyramid with base side B = 30 mm and height H = 60 mm (Step 213.3).

Left (Simple position):

Same as step 213.2, showing front view triangle (height 60 mm, base 30 mm) and top view square (30 mm × 30 mm).

Right (Tilted position):

Front view: Tilted triangle with face b'(c')-o' on the xy line, apex a'(d') at top.

Top view: Projected downward from front view and horizontally from initial top view. Corners of base: b (bottom-left), a (bottom-middle), c (top-left), d (top-middle). Apex o is at the far right. Visible edges include solid outlines b-c-d-o-a-b, while internal edges from d and a to o are shown as hidden (dashed) lines, and line b-c forms the base boundary.

Labels and annotations: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm', label '213.3', and parameter box 'B = 30', 'H = 60'.

Stage 3 Top View (Base edge at $30^\circ$ to VP): Draw a construction line below $xy$ inclined at $30^\circ$ to $xy$ , and reproduce the Stage 2 top view so the resting base edge $bc$ lies along this $30^\circ$ line.

Figure: Projection of a square pyramid - intermediate step 213.4

Orthographic projection step 213.4 of a square pyramid with base $B = 30\text{ mm}$ and height $H = 60\text{ mm}$ :

Stage 1 (Simple position):

Front View: Isosceles triangle with base of width $30\text{ mm}$ on the $xy$ reference line labeled $a'(d')$ on the left and $b'(c')$ on the right; apex $o'$ at height $60\text{ mm}$ .

Top View: A square of dimension $\square 30$ , corners labeled $a$ (bottom-left), $b$ (bottom-right), $c$ (top-right), $d$ (top-left), with diagonals intersecting at center $o$ .

Stage 2 (First tilt / lying on triangular face or slant edge on HP):

Front View: Triangular shape lying on the $xy$ line with apex $o'$ on the $xy$ line, base edge $b'(c')$ on $xy$ , and $a'(d')$ vertically above $b'(c')$ .

Top View: Apex $o$ projected to the right; base corners labeled $c, d$ at top and $b, a$ at bottom, connecting to apex $o$ .

Stage 3 (Second tilt in top view):

Top View reproduced and inclined such that the base edge line makes an angle of $30^\circ$ with the horizontal ( $xy$ reference line), with corners $b, c, d, a$ and apex $o$ pointing downwards to the right.

The final front view is not yet constructed in this step.

Stage 3 Front View: Project vertical lines up from all corners of the Stage 3 top view to meet horizontal locus lines from the Stage 2 front view. Connect the base $a'b'c'd'$ and slant edges to $o'$ , keeping edges from the back corner $b'$ dashed.

Figure: Projection of a square pyramid - completed solution 213.5

Complete orthographic projection 213.5 of a square pyramid with base $B = 30\text{ mm}$ and height $H = 60\text{ mm}$ :

Stage 1: Front view triangle (height $60$ , base $30$ on $xy$ ) and top view square (side $30$ , corners $a, b, c, d$ , center $o$ ).

Stage 2: Front view tilted with apex $o'$ and base point $b'(c')$ on the $xy$ line; corresponding top view projected below.

Stage 3:

Top view inclined at $30^\circ$ to the $xy$ line.

Final Front View constructed by projecting upwards from the inclined top view and horizontally from the stage 2 front view: points $b', c'$ on the $xy$ line, $a', d'$ at the top base level, and apex $o'$ on the $xy$ line. Outlines $b'-a'-d'-o'-b'$ are solid; interior edges from $d'$ down to $c'$ are shown, with hidden/visible edge conventions.

Answer: The complete three-stage orthographic projections of the square pyramid lying on its triangular face $obc$ with base edge $bc$ at $30^\circ$ to VP are shown in Figure 213.5.

Practice: True Inclination to Both Planes (Locus and Profile Plane Problems on HP)

Up to this point, you rotated the Stage 2 top view directly by a given top-view angle or edge condition. Now we tackle problems where the true inclination of an already-tilted axis or slant edge to the Vertical Plane (VP) is given. Whenever a problem specifies the true angle of a foreshortened axis or slant edge with VP, you must first construct a locus triangle to find the enlarged apparent angle $\beta$ before rotating the Stage 2 top view.

The idea

When an axis or slant edge of true length $L$ is tilted to the Horizontal Plane (HP) by an angle $\theta$ in Stage 2, it is no longer parallel to HP. Its Stage 2 top view foreshortens to a shorter length $L_1$ . If you rotate this foreshortened line $L_1$ directly at the given true angle $\phi$ to the $xy$ line in Stage 3, its true 3D angle with VP will be wrong because the line slopes upward while turning.

To keep the true angle with VP equal to $\phi$ , the vertical distance of the moving endpoint in front of VP must match the distance set by the true length $L$ inclined at $\phi$ . We draw a line from a fixed point (such as the base center $p$ or resting corner $c$ ) at the true angle $\phi$ to $xy$ and mark the true length $L$ along it. Through that endpoint, we draw a horizontal locus line parallel to $xy$ . Next, with the same fixed point as center and the shorter Stage 2 top-view length $L_1$ as radius, we swing an arc to cut the locus line. The line joining the fixed point to this intersection gives the apparent angle $\beta$ (where $\beta > \phi$ ), which is the true orientation of the Stage 2 top view.

A special case occurs when the true angles with HP and VP are complementary inclinations ( $\theta + \phi = 90^\circ$ , such as $50^\circ + 40^\circ = 90^\circ$ or $45^\circ + 45^\circ = 90^\circ$ ). Here, the foreshortened top-view length $L_1 = L \cos\theta$ equals $L \sin\phi$ , which is the exact perpendicular distance to the locus line. The arc of radius $L_1$ touches the locus line tangentially straight down from the fixed point, making the apparent angle $\beta = 90^\circ$ and placing the axis or slant edge in a Profile Plane (PP) perpendicular to both HP and VP.

Worked example

Problem 56: Pentagonal Pyramid with Slant Edge Inclined to Both Planes Given: Pentagonal pyramid, base edge 30 mm, height 60 mm, resting on a base corner on HP with the slant edge containing that corner inclined at $50^\circ$ to HP and $30^\circ$ to VP. Find: Three-stage orthographic projections and the apparent angle $\beta$ of the slant edge in the top view.

Draw the Stage 1 top view as a regular pentagon $abcde$ of side 30 mm with left edge $ae$ perpendicular to $xy$ so corner $c$ points to the right, and project the Stage 1 front view of height 60 mm.

Figure: Step 56.1: Initial projection of pentagonal pyramid resting on base in HP

Step 56.1 of the solution for question 56 showing the initial orthographic projections of a pentagonal pyramid. Top view: pentagon with vertical edge ae on the left (dimension 30 mm) and corner c pointing horizontally to the right. Slant edges connect corners a, b, c, d, e to the central apex o. Front view above reference line xy: base lies on the xy line with points a'(e') on the left, b'(d') inside, and c' on the right. Apex o' is directly above the center, with a vertical height dimension of 60 mm. The outline is triangle o'-a'(e')-c' with edge o'-b'(d') drawn as a visible line. An inset box reads B = 30, H = 60. Bottom notes state 'SCALE 1:1' and 'ALL DIMENSIONS ARE IN mm'.

Rotate the Stage 1 front view about corner $c'$ on $xy$ until the right-hand slant edge $o'c'$ makes $50^\circ$ with $xy$ because this slant edge is in true length in Stage 1.

Figure: Projection of a pentagonal pyramid - Step 56.2: Initial projections and tilted front view

Step 56.2 of the projection of a pentagonal pyramid.

Given box:

Problem number: 56.2

Base side ( $B$ ): 30 mm

Height ( $H$ ): 60 mm

Initial Views:

Ground/reference line labeled $x$ at left and $y$ at right.

Top View: Regular pentagon with apex at center $o$ and base vertices $a, b, c, d, e$ . Edge $e-a$ is vertical on the left with dimension 30 mm. Apex $o$ is connected to all base vertices $a, b, c, d, e$ .

Front View: Triangular elevation directly projected from the top view. Base lies on the reference line with points $a'(e')$ , $b'(d')$ , and $c'$ . Apex is labeled $o'$ at a vertical height of 60 mm from the base. Slant edges connect $o'$ to $a'(e')$ , $b'(d')$ , and $c'$ .

Second Stage (Tilted Front View):

The pyramid is tilted such that it rests on base corner $c'$ on the reference line $xy$ .

Slant edge $o'c'$ is inclined at an angle of $50^\circ$ to the $xy$ line / horizontal plane.

Vertices labeled: apex $o'$ , base points $c'$ , $b'(d')$ , and $a'(e')$ with $e'$ in parentheses above $a'$ . The axis is drawn as a center line from $o'$ to the midpoint of the base.

Notes:

SCALE 1:1

ALL DIMENSIONS ARE IN mm

Project vertical lines down from the tilted front view to meet horizontal locus lines from the Stage 1 top view, completing the Stage 2 top view where slant edge $oc$ is foreshortened.

Figure: Projection of a pentagonal pyramid - Step 56.3: Complete projection with second stage top view

Step 56.3 showing the final top view of the inclined pentagonal pyramid.

Given box:

Problem number: 56.3

Base side ( $B$ ): 30 mm

Height ( $H$ ): 60 mm

First stage and tilted front view:

Identical to step 56.2, with the front view tilted at $50^\circ$ on corner $c'$ on the $xy$ line.

Second Stage Top View Construction:

Vertical projection lines are drawn downwards from all vertices of the tilted front view ( $o'$ , $c'$ , $b'(d')$ , $a'(e')$ ).

Horizontal projection lines are drawn across to the right from the corresponding vertices of the first top view ( $a, b, c, d, e, o$ ).

Resulting final top view:

Apex $o$ is located on the right.

Base vertices labeled $a, b, c, d, e$ .

Outer contour edges $o-b$ , $b-a$ , $a-e$ , $e-d$ , $d-o$ and internal visible edges $o-a$ , $o-e$ are drawn as solid continuous lines.

Base edges $b-c$ , $c-d$ and slant edge $o-c$ are hidden (underneath) and drawn as dashed lines meeting at vertex $c$ .

Notes:

SCALE 1:1

ALL DIMENSIONS ARE IN mm

In Stage 3, draw a construction line from $c$ at the true angle $\phi = 30^\circ$ to $xy$ , step off the true slant-edge length $L = o'c'$ along it to draw a horizontal locus line, and swing an arc of radius $L_1 = oc$ (from Stage 2) to cut the locus at $o$ for an apparent angle $\beta = 51^\circ$ ; then copy the Stage 2 top view onto line $oc$ .

Figure: Orthographic projection steps of a pentagonal pyramid (base side 30 mm, height 60 mm) showing stage 1, stage 2, and stage 3 top view

Projection of a pentagonal pyramid with base side B = 30 mm and axis height H = 60 mm.

Stage 1:

Front View (FV1): Triangle resting on reference line xy with apex o' at height 60 mm. Base points are labeled a'(e'), b'(d'), and c'. Slant edges o'-a'(e'), o'-b'(d'), o'-c' are shown, along with a vertical center axis line.

Top View (TV1): Regular pentagon with vertical edge a-e of length 30 mm on the left, vertices labeled a, b, c, d, e, and central apex o connected to all vertices.

Stage 2:

Front View (FV2): The pyramid is tilted with base corner c' on the xy line. The base line passing through c' is inclined at an angle of 50° to the horizontal reference line. Base points are labeled (e')a', b'(d'), and apex o'.

Top View (TV2): Projected from FV2 and TV1. Points a, b, c, d, e form the base outline, and apex o is to the right. Solid lines show visible edges o-b, o-c, o-d, while hidden edges are dashed.

Stage 3:

Top View (TV3): The top view from stage 2 is reproduced with its axis inclined at 30° to the xy line, showing an APPARENT ANGLE of 51° with the reference horizontal line o-o1. Vertices labeled a, b, c, d, e, and apex o.

Front View (FV3): Omitted in this problem step.

Project upward from the Stage 3 top view to meet horizontal locus lines from the Stage 2 front view, drawing the visible upper slant faces with continuous lines and the hidden edges meeting at bottom corner $c'$ as dashed lines.

Figure: Complete three-stage orthographic projection of a pentagonal pyramid inclined to both reference planes

Complete three-stage projection of a pentagonal pyramid with base B = 30 mm and height H = 60 mm.

Stage 1:

Front View (FV1): Isosceles triangle with base on xy line of height 60 mm, apex o', base points a'(e'), b'(d'), c'. Centerline shown along the axis.

Top View (TV1): Regular pentagon with base edge a-e = 30 mm vertical on the left, vertices a, b, c, d, e, and center apex o connected to all five corners.

Stage 2:

Front View (FV2): Tilted with corner c' on xy, base line inclined at 50° to the horizontal xy line. Labeled points: c', b'(d'), (e')a', and apex o'.

Top View (TV2): Projected downwards from FV2 and horizontally from TV1. Shows apex o to the right, visible outer edges and dashed hidden lines.

Stage 3:

Top View (TV3): TV2 rotated so the axis is inclined at true angle 30° to the VP, corresponding to an APPARENT ANGLE of 51° to the horizontal reference line o-o1.

Front View (FV3): Final front view projected vertically from TV3 and horizontally from FV2. Vertices a', b', c', d', e' and apex o' are connected, with visible edges shown as solid lines and hidden edges as dashed lines.

Answer: Apparent angle $\beta = 51^\circ$ .

Practice: Solids Resting on the Vertical Plane (VP)

Having mastered every type of constraint for solids resting on HP, we now invert our workspace to solve problems where the solid rests with a point, edge, generator, or face on the Vertical Plane (VP). When a solid rests on VP, the entire three-stage procedure is mirrored across the XY line: draw the true base in the Stage 1 front view, tilt to VP in the Stage 2 top view, and rotate to HP in the Stage 3 front view.

The idea

When a solid rests on the Vertical Plane (VP), its axis is initially perpendicular to VP rather than HP. Because the base faces the observer directly in this simple position, the Stage 1 front view (above XY) shows the true shape of the base, while the Stage 1 top view (below XY) shows the solid's altitude with its resting base lying flat along the XY line. As always, orient the true-shape base in the front view so that the specific resting edge, corner, or base point lies at the extreme right.

In Stage 2, we introduce the inclination to VP ( $Ά$ ). Since distances in front of VP are seen in the top view, we rotate the Stage 1 top view below the XY line while keeping the resting point or edge on XY. Projecting upward from this tilted top view and across from the Stage 1 front view gives the foreshortened Stage 2 front view.

In Stage 3, we incline the solid to HP ( $θ$ ). Because inclination to HP is visible above the XY line, we rotate the Stage 2 front view directly by the given angle—or use a locus construction if a true angle is specified for an already-foreshortened axis. If the problem states that a corner or apex also touches HP, place that point directly on the XY line during this Stage 3 rotation. Finally, project downward to meet horizontal projectors from the Stage 2 top view to complete the final top view.

Worked example

Problem 81: A square prism of base side 40 mm and length 80 mm has a base edge on VP, axis inclined at 30° to VP and the resting base edge is inclined at 40° to HP. Draw the projections of the solid.

Given: Square prism, base side $B = 40$ mm, length $H = 80$ mm, resting on a base edge on VP, axis at 30° to VP, resting base edge at 40° to HP. Find: The three-stage orthographic projections of the prism.

Begin in the simple position with the axis perpendicular to VP so the base is parallel to VP and shows its true shape in the front view. Draw a 40 mm square above the XY line with its vertical edges perpendicular to XY, placing the resting base edge $2'3'$ on the right. Project downward to draw the Stage 1 top view as a 40 mm by 80 mm rectangle with the base $1(4) ext{--}2(3)$ resting directly on the XY line.

Figure: Step 81.1: Initial projection of square prism with base on VP

Step 81.1 shows the initial projections of a square prism:

Horizontal reference line labelled 'x' on the left and 'y' on the right.

Front view (above xy): A square of side 40 mm with vertices labelled top-left a'(1'), top-right b'(2'), bottom-right c'(3'), and bottom-left d'(4'). The top side is dimensioned as '□40'.

Top view (below xy): A rectangle of width 40 mm and length 80 mm aligned vertically with the front view. The top base edge lies on the xy line with vertices labelled 1(4) at the left and 2(3) at the right. The lower base edge has vertices labelled a(d) at the left and b(c) at the right. The axial length is dimensioned as 80 mm on the right.

A given box to the left contains: '81.1', 'B = 40', 'H = 80'.

Bottom text: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

Incline the axis at 30° to VP in the top view, where angles with VP appear in their true size. Reproduce the Stage 1 top view rectangle below XY with corner $2(3)$ on the XY line and the axis inclined at 30° to XY (so the base makes 60° with XY).

Figure: Orthographic projection of a square prism - Stage 1 and Stage 2 top view rotation

Step 81.2 of the projection of a square prism. GIVEN box: B = 40, H = 80. Reference line XY labeled with 'x' on the left and 'y' on the right. Stage 1 (Simple position):

Front view (above XY): Square of side 40 mm (dimensioned '□40' at top). Top-left corner labeled a'(1'), top-right b'(2'), bottom-right c'(3'), bottom-left d'(4').

Top view (below XY): Rectangle of width 40 mm and length 80 mm (dimensioned '80' on the right side). Top edge lies on XY line with left vertex labeled 1(4) and right vertex labeled 2(3). Bottom vertices labeled a(d) on left and b(c) on right. Stage 2:

Top view is reproduced rotated such that vertex 2(3) lies on the XY line and the long edge 2(3)-b(c) is inclined at an angle of 30° to the XY line.

Corners of the inclined top view labeled 2(3) at XY, b(c) at bottom-right, a(d) at bottom, and 1(4) at left.

Note at bottom: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

Project upward from all corners of the Stage 2 top view to intersect horizontal projectors from the Stage 1 front view. Connect the vertices to complete the Stage 2 front view, drawing the hidden resting edge $2'3'$ as a dashed line because it lies at the back (closest to VP).

Figure: Completed orthographic projection of a square prism inclined to VP

Step 81.3 of the projection of a square prism showing final front view projection. GIVEN box: B = 40, H = 80. Stage 1:

Same as in step 81.2 (square front view of side 40 mm, rectangular top view of 40x80 mm with top edge on XY). Stage 2:

Top view: Rectangle 40x80 mm inclined at 30° to the XY line with corner 2(3) touching the XY line. Corners labeled 2(3), b(c), a(d), 1(4).

Front view: Projected vertically upwards from the rotated top view and horizontally from the first stage front view.

Top horizontal line contains vertices: 1', 2', a', b'.

Bottom horizontal line contains vertices: 4', 3', d', c'.

Outer boundary 1'-b'-c'-4' drawn with solid lines.

Vertical edge a'-d' is a continuous visible solid line.

Vertical edge 2'-3' is a dashed hidden line.

Note at bottom: 'SCALE 1:1', 'ALL DIMENSIONS ARE IN mm'.

Since the resting base edge $2'3'$ remains in true length on VP, rotate the entire Stage 2 front view directly so that edge $2'3'$ is inclined at 40° to the XY line.

Figure: Projection of a square prism, Step 81.4

Step 81.4 in orthographic projection of a square prism.

Given box: B = 40, H = 80.

Scale: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

Stage 1 (Initial position):

Front View (above reference line): Square of dimension □40, vertices labeled top-left a'(1'), top-right b'(2'), bottom-right c'(3'), bottom-left d'(4').

Top View (below reference line): Rectangle 40 mm wide and 80 mm long, vertices labeled top-left 1(4), top-right 2(3), bottom-left a(d), bottom-right b(c). Height dimension 80 marked.

Stage 2 (First tilt):

Top view rotated such that long edge 2(3)–b(c) is inclined at 30° to the horizontal reference line.

Front view projected onto reference line xy: base line on xy from 4' to c' passing through 3' and d', top line from 1' to b' passing through 2' and a'. A vertical dashed line connects 2' to 3'.

Stage 3 (Second tilt - incomplete):

Front view reproduced and tilted so that the base line 4'-3' makes an angle of 40° with the horizontal line.

Vertices labeled 1', 2', 3', 4', a', b', c', d'. Top view is yet to be drawn.

Project vertically downward from every corner of the Stage 3 front view to intersect horizontal projectors from the Stage 2 top view, and join the points with visible and hidden lines to complete the final top view.

Figure: Projection of a square prism, Step 81.5 (Completed projection)

Step 81.5 showing the final stage of the orthographic projection of the square prism.

Given box: B = 40, H = 80.

Scale: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

Stages 1 and 2 are identical to Step 81.4.

Stage 3 (Final projection):

Front view: Tilted at 40° to xy line with vertices 1', 2', 3', 4', a', b', c', d'.

Top view: Completed by projecting downward from the Stage 3 front view and horizontally across from the Stage 2 top view. Vertices labeled 1, 2, 3, 4, a, b, c, d.

Outlines and visible edges drawn solid: 2-3-c-d-a-1-2, 3-b, b-c, 2-b.

Hidden edges drawn dashed: connecting to vertex 4 (edges 1-4, 3-4, and 4-d).

Answer: Completed three-stage projections of the 40 mm by 80 mm square prism with its axis at 30° to VP and resting base edge at 40° to HP.

Practice: Real-World Equilibrium—Freely Suspended Pyramids

You already know how to tilt solids through three stages and how to construct a locus triangle when an already-inclined axis makes a true angle with VP. Now we combine those geometric skills with physics to predict how a three-dimensional pyramid hangs in space when freely suspended from a corner by a string. A pyramid suspended from a base corner naturally settles so that its center of gravity (one-quarter of the height up the axis) hangs directly below the suspension corner.

The idea

When you tie a string to one base corner of a solid and let it hang freely, gravity pulls the solid's entire weight downward through a single balance point called the center of gravity ( $G$ , front view $g'$ ). The solid swings until it reaches physical equilibrium, which happens only when $g'$ comes to rest vertically below the suspension corner. In a front view, "vertically below" means the line joining the suspension corner and $g'$ is perpendicular to the $xy$ line.

For any pyramid or cone of height $H$ , the center of gravity always lies on the central axis at a distance of $H/4$ from the base. For example, if the axis height is 60 mm, you measure $60/4 = 15$ mm up the axis from the base in the Stage 1 front view to mark $g'$ . You must also orient the Stage 1 top view so that the chosen suspension corner lies at an extreme left or right position, ensuring the line from that corner to $g'$ shows its true relationship in the front view.

Instead of being given a numerical angle with HP for Stage 2, the suspension rule sets the tilt automatically: redraw the Stage 1 front view so the line connecting the suspension corner to $g'$ is perpendicular to $xy$ (with the apex $o'$ resting on $xy$ for convenience). Finally, complete Stage 3 either by rotating the top view of the axis directly if the apparent angle with $xy$ is given, or by constructing an apparent-angle locus if the true inclination to VP is specified.

Worked example

Problem 64: Freely suspended triangular pyramid (top view inclined)

Given: A triangular pyramid of base edge 40 mm and axis 60 mm is freely suspended from a base corner, with the top view of its axis inclined at 50° to the $xy$ line. Find: Its three-stage front and top views.

Draw the Stage 1 top view as an equilateral triangle $abc$ of side 40 mm with edge $bc$ perpendicular to $xy$ on the right, placing the suspension corner $a$ at the extreme left so it tilts cleanly in the front view.
Project up to draw the Stage 1 front view with base $a'b'(c')$ on $xy$ and apex $o'$ at height 60 mm, then mark the center of gravity $g'$ on the axis at $H/4 = 60/4 = 15$ mm above the base and join $a'g'$ .

Figure: Projection of triangular pyramid in simple position (Step 64.1)

Step 64.1 of the projection of a triangular pyramid:

A given box specifies: 'B = 40', 'H = 60'.

Front view (above the xy-line): An isosceles triangle with base corners a' and b'(c') resting on the xy reference line. Base edge a'b'(c') is horizontal. The apex is labeled o' at a vertical height of 60 mm from the base. A dashed centerline extends vertically from the base through the center of gravity g' (located at height 60/4 = 15 mm above the base) to the apex o'.

Top view (below the xy-line): An equilateral triangle with corners a on the left, and vertical edge bc on the right (c at top, b at bottom). Base edge ab has dimension 40 mm. Apex o is at the centroid, with slant edges connecting o to a, o to b, and o to c.

Vertical projector lines connect corners a to a', b and c to b'(c'), and apex o to axis line/o'.

For Stage 2, redraw the front view so the suspension line $a'g'$ is perpendicular to $xy$ (resting apex $o'$ on $xy$ ), because a freely hanging body settles with its center of gravity directly below the point of suspension.

Figure: Orthographic projection step 64.2 of a triangular pyramid

Step 64.2 showing the initial orthographic projections and the tilted front view of a triangular pyramid.

GIVEN box: B = 40, H = 60.

Notes: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

Stage 1 (Initial position):

Reference line labeled x-y.

Front view: An isosceles triangle with base on the xy line from a' to b'(c'). Total height is 60 mm. The center of gravity g' is marked on the central axis at a height of 60/4 = 15 mm above the base.

Top view: Equilateral triangle with base edge bc vertical on the right (length 40 mm) and vertex a pointing to the left. The apex o is at the center, connected to vertices a, b, and c by slant edges.

Stage 2 (Tilted front view):

The pyramid's front view is redrawn tilted to the right, showing base vertices b'(c') and a', apex o', and center of gravity g' with a dashed line connected towards o'.

Project downward from the tilted Stage 2 front view and across from the Stage 1 top view to complete the Stage 2 top view, marking the foreshortened axis and hidden edges.

Figure: Orthographic projection step 64.3 of a triangular pyramid

Step 64.3 showing the completed two-stage projection of the triangular pyramid.

GIVEN box: B = 40, H = 60.

Notes: SCALE 1:1, ALL DIMENSIONS ARE IN mm.

Stage 1 (Initial position):

Front view: Base a'-b'(c') on the xy line, height 60 mm, center of gravity g' at height 60/4 = 15 mm.

Top view: Triangle abc with base side 40 mm, apex o at the center connected to vertices a, b, and c.

Stage 2 (Tilted position):

Front view: Re-drawn from step 64.2 with apex o' tilted to the right, base edge b'(c')-a', and center of gravity g'.

Top view: Obtained by projecting vertically downwards from front view vertices a', b'(c'), o' and horizontally from the first stage top view vertices a, b, c, o. Base vertices b and c remain on vertical alignment, apex o is on the far right, and apex lines connect to all vertices.

Since the problem gives the angle of the top view of the axis directly, draw a line inclined at 50° to $xy$ and reproduce the Stage 2 top view along it to get the Stage 3 top view.

Figure: Orthographic projection steps of a triangular pyramid - Step 64.4

Step 64.4 showing projection of a triangular pyramid with base edge B = 40 mm and height H = 60 mm.

First stage:

Front view shows an isosceles triangle on reference line x-y with base points a' to b'(c') = 40 mm, apex at top with height 60 mm, and center of gravity g' marked at height 60/4 = 15 mm above base.

Top view directly below shows an equilateral triangle with vertices a, b, c and apex o at the center connected to vertices. Side length = 40 mm.

Second stage:

Front view is tilted so it rests on apex o' with base b'(c') elevated and apex a' at the highest point, with g' along the axis line.

Top view is projected below with vertices a, b, c, o.

Third stage:

Top view is reproduced rotated such that its edge/axis makes an angle of 50° with the horizontal/reference line. Note: Scale 1:1, all dimensions are in mm.

Project vertically up from the Stage 3 top view and horizontally across from the Stage 2 front view to complete the final Stage 3 front view, checking visibility from the bottom of the top view.

Figure: Completed orthographic projection of a triangular pyramid - Step 64.5

Step 64.5 showing the complete three-stage projection of the triangular pyramid (B = 40 mm, H = 60 mm).

First stage: Front view of height 60 mm with base on x-y line, center of gravity g' at 15 mm (60/4), and equilateral triangular top view of base edge 40 mm.

Second stage: Tilted front view resting on apex o' with base b'(c') tilted, and the corresponding top view projected downwards.

Third stage: Top view rotated at 50° to the horizontal/projector line, and projectors drawn vertically upward intersecting horizontal projectors from stage 2 to construct the final front view. In the final front view, visible edges connect apex a' to b', c', and o', while base edge b'-c' is drawn as a hidden dashed line. Note: Scale 1:1, all dimensions are in mm.

Answer: The pyramid hangs with line $a'g'$ vertical ( $g'$ at 15 mm from the base along the 60 mm axis) and its top-view axis rotated directly at 50° to $xy$ .

Orthographic Projections of Solids Inclined to Both Reference Planes

Your path through this module

Classification of Solids and the Simple Position

Worked example

Remember

The Three-Stage Change-of-Position Method and Visibility Rules

Worked example

Remember

Solids of Revolution and Resting on Faces, Slant Edges, or Generators

Worked example

Remember

True vs. Apparent Inclination (The Locus Method) and Profile Plane Cases

Worked example

Remember

Freely Suspended Solids and Solids Resting on the Vertical Plane (VP)

Worked example

Remember

Practice: Polyhedra on HP with Direct Stage-3 Rotation

The idea

Worked example

More problems

Problem Q51: Square pyramid, apex away from VP, right

Problem Q52 (53): Square pyramid, apex towards VP, right

Problem Q53 (54): Square pyramid, apex away from VP, left

Problem Q54 (55): Square pyramid, apex towards VP, left

Problem Q55 (56): Pentagonal pyramid on a base corner

Problem Q250 (59): Hexagonal prism with resting edge inclined

Problem Q79 (60): Hexagonal prism with axis top view inclined

Practice: Cones and Cylinders on HP with Direct Stage-3 Rotation

The idea

Worked example

More problems

Problem 70: Cone with Horizontal Base Diameter Inclined to VP

Problem 219: Cylinder Inclined to HP and Top View Inclined

Problem 71: Cone Lying on a Generator on HP

Problem 73: Cone on Generator on HP with Axis on Profile Plane

Problem 75: Cone Lying on a Generator on VP with Base Inclined to HP

Practice: Pyramids and Cubes on HP with Special Face, Edge, and Diagonal Conditions

The idea

Worked example

More problems

Problem 58: Hexagonal pyramid lying on a triangular face

Problem 59: Hexagonal pyramid lying on a lateral edge

Problem 60: Square pyramid with triangular face parallel to VP

Problem 197: Pentagonal pyramid with triangular face perpendicular to HP

Problem 61: Square pyramid with lateral edge perpendicular to HP

Problem 62: Square pyramid resting on apex with vertical slant edge

Problem 82: Cube with body diagonal perpendicular to VP

Practice: True Inclination to Both Planes (Locus and Profile Plane Problems on HP)

The idea

Worked example

More problems

Problem 80: Hexagonal Prism with Axis Inclined to HP and VP

Problem 85: Cylinder Resting on Base Point with Axis Inclined to Both Planes

Problem 72: Cone Lying on a Generator on HP with Axis Inclined to VP

Problem 57: Pentagonal Pyramid with Slant Edge in a Profile Plane

Problem 84: Pentagonal Prism with Axis in a Profile Plane

Practice: Solids Resting on the Vertical Plane (VP)

The idea

Worked example

More problems

Problem 74: Cone resting on base point on VP

Problem 63: Hexagonal pyramid with triangular face on VP

Problem 76: Cone resting on generator on VP

Problem 198: Cone on base point on VP (locus)

Practice: Real-World Equilibrium—Freely Suspended Pyramids

The idea

Worked example

More problems

Problem 65: Freely suspended square pyramid (locus method)